Question: Evaluate the following expression. Your answer must be exact. $\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)^9=$
Explanation: The Strategy The easiest way to find $z^{n}$ for a complex number $z=({a}+{b}i)$ is using its modulus and argument. Therefore, our solution will consist of the following steps: Find the modulus and argument of $z$. [How is this done, in general?] Find the modulus and argument of $z^{n}$. [How is this done, in general?] Find the rectangular form $z^{n}$. Find the modulus and argument of $\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)$ $\left({-\dfrac{\sqrt{3}}{2}}+{\dfrac{1}{2}}i\right)$ is of the form $({a}+{b}i)$, where ${a=-\dfrac{\sqrt{3}}{2}}$ and ${b=\dfrac{1}{2}}$. Therefore: $\begin{aligned}r&=\sqrt{{a}^2 + {b}^2} \\\\&=\sqrt{ \left({-\dfrac{\sqrt{3}}{2}}\right)^2 + \left({\dfrac{1}{2}}\right)^2} \\\\&=\sqrt{{\dfrac{3}{4}}+{\dfrac{1}{4}}} \\\\&=1\end{aligned}$ Using the arctangent formula, we have: $\begin{aligned}\theta&=\arctan\left(\dfrac{{b}}{{a}}\right) \\\\&=\arctan\left(\dfrac{{\dfrac{1}{2}}}{{-\dfrac{\sqrt{3}}{2}}}\right) \\\\&=-30^\circ\end{aligned}$ Since ${a=-\dfrac{\sqrt{3}}{2}}$ is negative and ${b=\dfrac{1}{2}}$ is positive, $\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)$ lies in Quadrant $2$. Therefore, $\theta$ must be between $90^\circ$ and $180^\circ$. Using the identity $\tan(180+\theta)=\tan(\theta)$, we know that the following is also a solution of the equation. $180^\circ+(-30^\circ)=150^\circ$ So $\theta = 150^{\circ}$. Find the modulus and argument of $\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)^9$ We found that the modulus and argument of $\left({-\dfrac{\sqrt{3}}{2}}+{\dfrac{1}{2}}i\right)$ are $1$ and $150^\circ$. Therefore, the modulus and argument of $\left({-\dfrac{\sqrt{3}}{2}}+{\dfrac{1}{2}}i\right)^9$ are $1^9=1$ and $(150^\circ)\cdot9=1350^\circ=270^\circ$. Find the rectangular form of $\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)^9$ Since the argument is $270°$, we know the number lies on the negative side of the imaginary number axis and is therefore a negative pure imaginary number. Since the modulus is $1$, our solution is $-i$. [What does this look like graphically?] [How do we find this algebraically?] Summary $\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)^9=-i$